In the coordinate plane, let AB be a segment of length 102, and consider the circle with diameter AB. Points P and Q lie on this circle and satisfy (PA+PB)⋅(PQ+PB)=2PQ2. If PB=14 and QB>0, then PA⋅QB=pq, where p and q are coprime natural numbers. Compute p+q. [4 points]
Ground Truth
221
Execution Trace
Reasoning Process
[THOUGHT]
To solve the problem, we start by placing the circle with diameter AB on the coordinate plane. Let B be at the origin (0,0) and A at (102,0). The radius of the circle is 52, and the center is at (52,0).
Given ∣PB∣=14, we determine the coordinates of P. Since P lies on the circle, it satisfies the equation:
(xp−52)2+yp2=50
and the distance from P to B is:
xp2+yp2=14⟹xp2+yp2=196
Solving these equations, we find:
xp=5492,yp=±572
Next, we use the given vector equation:
(PA+PB)⋅(PQ+PB)=2∣PQ∣2
Expressing the vectors in terms of coordinates, we get: